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J. Math. Anal. Appl. 377 (2011) 710–731

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Journal of Mathematical Analysis andApplications

www.elsevier.com/locate/jmaa

Concerning the regularity of the anisotropic porous medium equation ✩

Eurica Henriques

Departamento de Matemática, Edifício das Ciências Florestais, Quinta dos Prados, Apartado 1013, 5001-801 Vila Real, Portugal

a r t i c l e i n f o a b s t r a c t

Article history:Received 29 April 2010Available online 10 November 2010Submitted by Goong Chen

Keywords:Anisotropic porous medium equationDegenerate/singular pdesRegularity theoryIntrinsic scaling

We show that a locally bounded nonnegative weak solution of the anisotropic porousmedia equation is locally continuous. The proof is based on DiBenedetto’s techniquecalled intrinsic scaling; by choosing an appropriate geometry one can deduce energy andlogarithmic estimates from which one can implement an iterative method to obtain theregularity result.

© 2010 Elsevier Inc. All rights reserved.

1. Introduction

The study of nonlinear partial differential equations gained a significant importance in the recent past years, not only fortheir physical and biological relevance but, and no less important, also for the mathematical novelties intrinsically relatedto the subject. The development of the regularity theory for degenerate and/or singular parabolic pdes is one exampleof the contemporary analysis of nonlinear pdes. One has to go back to the final fifties to encounter the now standardprocedure that allows one to get a regularity result for the solutions of nonlinear pdes: regularity for linear elliptic pdeswas established by De Giorgi [8]; while Moser [13,15,14,16] and Nash [17] dealt with linear parabolic pdes. Afterwards theRussian school generalized De Giorgi’s proof for quasi-linear elliptic equations and also for quasi-linear parabolic equationsthat possess a linear growth with respect to the modulus of the gradient of the solution, |∇u|. Following the main andprincipal ideas of De Giorgi, DiBenedetto [6,7] developed an interesting and powerful procedure that allowed to fill the gapleft open by the Russian school. DiBenedetto realized that the “bad” structure of the quasi-linear parabolic pdes should betaken into account; the method of intrinsic scaling was a remarkable breakthrough on the regularity issue for those typeof pdes. By incorporating the singularity/degeneracy of the equation in a proper geometry, it is possible to derive integralinequalities, obtained for the “right” cylinders, such that one can heuristically say that the pde behaves, in its own specificgeometry, as the heat equation ut − �u = 0.

Regarding to the porous medium equation (pme)

β(u)t − �u = 0, β(u) = |u| 1m sign u, m > 1, (1.1)the regularity theory was addressed from the final sixties to the eighties by several authors such as Aronson, Caffarelli,Friedman, Evans, Sacks, Ziemer and DiBenedetto. Aronson [1] proved that the weak solutions of (1.1) defined in R × R+are locally Hölder continuous with respect to the space variable; the time regularity was done separately by Gilding [9],Kružkov [12] and DiBenedetto [4]. Caffarelli and Friedman [2] obtained the continuity and the Hölder continuity of the

✩ Research supported by CM-UTAD/FCT and Project PTDC/MAT/098060/2008. The author would like to thank Professor Shmarev for all the fruitfuldiscussions on the subject.

E-mail address: [email protected].

0022-247X/$ – see front matter © 2010 Elsevier Inc. All rights reserved.doi:10.1016/j.jmaa.2010.10.077

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E. Henriques / J. Math. Anal. Appl. 377 (2011) 710–731 711

nonnegative solutions of (1.1) defined in ΩT ⊂ RN+1, N > 1. As for the pme with a relaxation of the sign of u, Caffarelli andEvans [3] proved the continuity of the weak solutions of (1.1) and, using methods close to these ones, Sacks [18] obtainedthe continuity of the solution of (1.1) (incorporating within the same proof the pme, the fast diffusion equation and also thetwo phase Stefan problem). Ziemer [21] and DiBenedetto [5] extended the previous results in the sense that they studiedthe more general equation

β(u)t − ∇ · a(x, t, u,∇u) + b(x, t, u,∇u) = 0,considering suitable assumptions on a and b. The proofs followed different approaches: while DiBenedetto’s proof was basedon a parabolic version of De Giorgi’s technique, Ziemer’s approach was related to the Moser’s iteration technique.

Recently more generalizations of the pme were considered. In [11] and [10], Henriques and Urbano proved that the localweak solutions of

ut − ∇ ·(uγ (x,t)∇u) = 0

are locally continuous, assuming the function γ (x, t) to be a bounded function −1 < γ − � γ (x, t) � γ + < ∞ and satisfyingthe regularity assumption γ ∈ L∞(0, T ; W 1,p(Ω)), for some p > max{2, N}. The nonstandard growth condition exhibited bythis equation played a significant role: the x-dependence on γ ultimately implied that the constants appearing along theproof present a dependence on the oscillation of the solution itself. Therefore the authors were unable to derive an interiormodulus of continuity for u.

In this paper we are concerned with the anisotropic pme modeled by

ut −N∑

i=1

(umi

)xi xi

= 0, in ΩT = Ω × (0, T ], 0 < T < ∞, Ω ⊂ RN . (1.2)

Our aim is to establish an interior regularity result for the solutions of (1.2). We will adapt DiBenedetto’s technique to provethat a local weak solution u of (1.2) is locally continuous, assuming u to be a nonnegative bounded function and all theexponents mi to be positive real numbers. The degenerate and the singular cases are treated separately but in an unifyingway, as usual. By studying an alternative argument, we will show that for each point (x0, t0) ∈ int ΩT we can constructa sequence of nested and shrinking cylinders Q n and a sequence of oscillations ωn such that ωn → 0 as Q n → (x0, t0).Along the proof for both the degenerate and singular cases, Section 3 and Section 4, respectively, one obtains constantsthat depend on the oscillation itself (see for instance Remark 5, Remark 9 and Remark 10). Consequently one can no longerexpect to present a modulus of interior continuity for the solution. The Hölder continuity is lost.

To the best of the author’s knowledge, no regularity result is known for the anisotropic case; not even in the context of“very weak” solution presented by Song in [19,20], where he obtained results concerning the existence and the uniquenessof the “very weak” solution of the anisotropic pme with singular advections and absorptions.

2. Regularity result

Consider the anisotropic porous medium equation

ut −N∑

i=1

(umi

)xi xi

= 0, in ΩT , (2.1)

where ΩT = Ω × (0, T ] is a bounded domain in RN+1, 0 < T < ∞, and 0 < m � mi � m < ∞, ∀i, beingm = min

1�i�Nmi, m = max

1�i�Nmi .

Clearly that Eq. (2.1) presents a degradation of its parabolic structure: if m > 1, (2.1) is a degenerate pde since the diffusioncoefficient umi−1 is zero at the points where u = 0; if by the contrary 0 < m < 1, we are in the presence of a singularparabolic pde since the diffusion coefficient explodes at the points where u = 0. Both cases will be treated in an unifyingway, that is, we will show that for each interior point (x0, t0) of ΩT one can construct a sequence of nested and shrinkingcylinders Q n , all of them with vertex at (x0, t0), and a sequence of real positive numbers ωn , such that ωn goes to zero asthe cylinders shrink to their vertex. Namely we will prove,

Theorem 1. Any locally bounded nonnegative weak solution of (2.1) is locally continuous in ΩT .

First, we have to start by assuming that u is a nonnegative bounded function satisfying (2.1) and then we define whatwe mean by a locally weak solution of Eq. (2.1).

Definition 1. We say that a measurable function u is a locally weak solution of (2.1) if

712 E. Henriques / J. Math. Anal. Appl. 377 (2011) 710–731

• u ∈ L∞(0, T ; L∞(Ω)) with u(x, t) ∈ [0,1] a.e. in ΩT ;• u ∈ C(0, T ; L2(Ω)) and u mi−12 uxi ∈ L2(0, T ; L2(Ω)), i = 1, . . . , N;• for every compact K ⊂ Ω and for every subinterval [t1, t2] ⊂ (0, T ],

∫K

uφ

∣∣∣∣t2

t1

+t2∫

t1

∫K

{−uφt +

∑(umi

)xiφxi

}= 0, (2.2)

for all functions φ ∈ H1loc(0, T ; L2(K )) ∩ L2loc(0, T ; H10(K )).

Remark 1. Observe that the definition of weak solution does not depend neither on the degeneracy nor on the singularityof the anisotropic pme.

By using the Steklov average of a function (see [6] for more details) (2.2) is equivalent to the following formulation

• for every compact K ⊂ Ω and for all 0 < t < T − h,∫K×{t}

(uh)tφ dx +∑ ∫

K×{t}

(miu

mi−1uxi)

hφxi dx = 0, (2.3)

for all φ ∈ H10(K ).

Remark 2. Parabolic equations possess a low degree of regularity regarding the time variable and thereby, to the accurate,one should work with (2.3) instead of (2.2). However, for the sake of simplicity the author decided to perform formalcalculations along the proofs.

In this paper we will present a fully detailed proof concerning the local continuity of a local weak solution of thedegenerate equation (2.1); as for the singular case we will briefly comment on the changes to be performed since the proofis essentially the same. The constants appearing in the proof usually will be denoted by C(N,m,m ), this means that theconstants depend on the parameters in brackets. We stress the fact that the constants may vary from line to line.

In terms of notation, being

Q = K R × (t0, t1) ={

x: max1�i�N

|xi | < R}

× (t0, t1)

a parabolic cylinder with hedge 2R and height t1 − t0, its parabolic boundary is represented by∂p Q :=

(∂ K R × [t0, t1]

) ∪ (K R × {t0}).3. The degenerate equation

Assuming that m > 1, Eq. (2.1) is a degenerate pde.In order to obtain the interior continuity of the solutions by means of intrinsic scaling, we need to consider a geometry

that accommodates the degeneracy of the anisotropic pme (2.1). For that purpose let (x0, t0) be an interior point of thespace–time domain ΩT that, by translation, we may assume to be (0,0). Consider small positive numbers > 0 and R > 0such that the cylinder

Q(

R2−, R) := K R × (−R2−,0)

is a subset of ΩT and define

μ− := ess infQ (R2− ,R)

u; μ+ := ess supQ (R2− ,R)

u; ω := ess oscQ (R2− ,R)

u = μ+ − μ−.

Construct the cylinder

Q(a0 R

2, R), a0 =

(ω

4

)1−m

1

and assume that

ω � 4R

m−1 . (3.1)


This assumption implies

Q(a0 R

2, R) ⊂ Q (R2−, R)

and the following relation between the oscillations

ess oscQ (a0 R2,R)

u � ω. (3.2)

Remark 3. If (3.1) does not hold then there is nothing else to be done since the essential oscillation ω goes to zero as theradius R goes to zero.

Recall that, in general, (3.2) does not hold for a given cylinder; the cylinder dimensions must be defined in terms of theessential oscillation within it.

If m = m = 1, a0 = 1 and we recover the standard parabolic cylinder with the natural homogeneity of the space and timevariables.

The degenerate nature of (2.1) is accommodated in the chosen geometry Q (a0 R2, R). Within this geometry it is possibleto obtain, from the energy and logarithmic estimates, integral norms which allow us to heuristically see (2.1) almost as theheat equation. Even so, since the constants appearing along the proof depend on ω, one cannot explicitly derive an interiormodulus of continuity for u.

The technique of intrinsic scaling is implemented by working with an alternative argument.Given ν0 ∈ (0,1), to be determined in terms of the data and ω, either

the first alternative∣∣∣∣(x, t) ∈ Q (a0 R2, R): u(x, t) < ω2∣∣∣∣ � ν0∣∣Q (a0 R2, R)∣∣ (3.3)

or

the second alternative∣∣∣∣(x, t) ∈ Q (a0 R2, R): u(x, t) > μ+ − ω2∣∣∣∣ < (1 − ν0)∣∣Q (a0 R2, R)∣∣. (3.4)

Remark 4. In the first and second alternatives we assumed that μ− = 0. This assumption is easily justified by the fact that0 � u � 1 and that Eq. (2.1) has a degradation of its structure at the points where u = 0.

Combining the results of the two alternatives we will be able to obtain a reduction of the oscillation of the solutionwithin a smaller cylinder. Namely

Proposition 1. There exist positive numbers ν0, σ (ω) ∈ (0,1), depending on the data and on ω, such thatess osc

Q (ν02 a0(

R2 )

2, R2 )

u � σ(ω)ω. (3.5)

3.1. Reduction of the oscillation when u is essentially away from zero

In this section we analyze the first alternative. We will obtain explicitly the value ν0, which will depend on ω, and wewill show that within the smaller cylinder Q (a0( R2 )

2, R2 ) the essential oscillation decreases by the factor34 .

Proposition 2. There exists ν0 ∈ (0,1), depending on the data and ω, such that if (3.3) holds true then

u(x, t) � ω4

, a.e. (x, t) ∈ Q(

a0

(R

2

)2,

R

2

). (3.6)

Proof. Define two decreasing sequences of positive numbers

Rn = R2

+ R2n+1

, kn = ω4

+ ω2n+2

, n = 0,1, . . .and construct the family of nested and shrinking cylinders Q (a0 R2n, Rn).


Introduce the function uω = max{u, ω4 }. In the weak formulation (2.2) take φ = −(uω − kn)−ξ2n , where 0 � ξn � 1 aresmooth cutoff functions defined in Q n = Q (a0 R2n, Rn) and satisfying⎧⎪⎨

⎪⎩ξn ≡ 1 in Q n+1, ξn ≡ 0 on ∂p Q n;

|∇ξn| � 2n+2

R, |�ξn| � 2

2(n+2)

R2, 0 < ξnt �

22(n+2)

a0 R2,

and consider the integration in time over (−a0 R2n, t), for t ∈ (−a0 R2n,0). We obtain

I1 + I2 :=t∫

−a0 R2n

∫K Rn

ut[−(uω − kn)−ξ2n ] + ∑

t∫−a0 R2n

∫K Rn

(umi

)xi

[−(uω − kn)−ξ2n ]xi= 0.

Observe that

I1 = 12

t∫−a0 R2n

∫K Rn

∂t(uω − kn)2−ξ2n +(

ω

2n+2

) t∫−a0 R2n

∫K Rn

−utξ2n χ[u< ω4 ]

= 12

∫K Rn ×{t}

(uω − kn)2−ξ2n −t∫

−a0 R2n

∫K Rn

(uω − kn)2−ξnξnt +(

ω

2n+2

) ∫K Rn×{t}

(u − ω

4

)−ξ2n

− 2(

ω

2n+2

) t∫−a0 R2n

∫K Rn

(u − ω

4

)−ξnξnt

� 12

∫K Rn ×{t}

(uω − kn)2−ξ2n − 3(

ω

4

)2 22(n+2)a0 R2

t∫−a0 R2n

∫K Rn

χ[uω�kn],

since the third term is nonnegative and, for 0 � u � ω4 ,

uω = ω4

� kn

and, for ω4 < u = uω � kn ,

(uω − kn)− � kn − uω = kn − u < kn − ω4

= ω2n+2

� ω4

.

As for the integral evolving the space derivatives we get

I2 =∑ t∫

−a0 R2n

∫K Rn

miumi−1ω

∣∣((uω − kn)−)xi ∣∣2ξ2n + 2∑t∫

−a0 R2n

∫K Rn

miumi−1ω

((uω − kn)−

)xiξn(uω − kn)−ξnxi

+ 2(

ω

2n+2

)∑ t∫−a0 R2n

∫K Rn

( ω4∫u

mi smi−1 ds

)xi

ξnξnxi χ[u< ω4 ]

� 12a0

m

t∫−a0 R2n

∫K Rn

∣∣∇(uω − kn)−∣∣2ξ2n − 2m(

ω

4

)2 22(n+2)R2

t∫−a0 R2n

∫K Rn

χ [uω < kn]

− 4(

ω

4

)2(ω

4

)m−1 22(n+2)R2

t∫−a0 R2n

∫K Rn

χ

[u <

ω

4

].

The above inequalities were obtained by Cauchy’s inequality with = 12 , integration by parts, the fact that for ω4 < u =uω � kn we get


(uω − kn)− = kn − uω = kn − u < ω2n+2

� ω4

;1

a0<

(ω

4

)mi−1< umi−1ω � 1;

and for u � ω4ω4∫

u

mi smi−1 ds =

(ω

4

)mi− umi �

(ω

4

)mi�

(ω

4

)m.

From the estimates obtained before we get the energy estimates

sup−a0 R2n


Taking

ν0 ≡ C(m, N)− N+22 2−(N+2)2ω(m−1) N+22 (3.7)we have

Y0 � C(m, N)−N+2

2 2−(N+2)2ω(m−1)N+2

2

and then, by Lemma 4.1 of [6, p. 12], we can conclude that Yn → 0 as n → ∞.Noting that Rn ↘ R2 , kn ↘ ω4 , and Yn → 0 as n → ∞ implies that An → 0 as n → ∞, we obtain∣∣∣∣(x, z) ∈ Q

((R

2

)2,

R

2

): ūω(x, z) �

ω

4

∣∣∣∣ = 0,that is,

u(x, t) � ω4

, a.e. (x, t) ∈ Q(

a0

(R

2

)2,

R

2

). �

We immediately get

Corollary 1. There exist constants ν0 ∈ (0,1), depending on the data and ω, and σ0 ∈ (0,1), such that if (3.3) holds theness osc

Q (a0(R2 )

2, R2 )

u � σ0ω. (3.8)

3.2. Reduction of the oscillation when u is essentially away from its supremum

Now we focus the study on the complementary case, that is, we now assume that the solution u is essentially awayfrom its supremum, i.e., (3.4) is in force.

Being the number ν0 already determined, we will deduce a result similar to Corollary 1. The proof is done in severalsteps. The first one is the following.

Lemma 1. Assume that (3.4) holds true. There exists a time level

t0 ∈[−a0 R2,−ν0

2a0 R

2]

(3.9)

such that∣∣∣∣x ∈ K R : u(x, t0) > μ+ − ω2∣∣∣∣ <

(1 − ν01 − ν02

)|K R |. (3.10)

Proof. In fact, if not

∣∣∣∣(x, t) ∈ Q (a0 R2, R): u(x, t) > μ+ − ω2∣∣∣∣ �

− ν02 a0 R2∫−a0 R2

∣∣∣∣x ∈ K R : u(x, t) > μ+ − ω2∣∣∣∣dt

�(

1 − ν01 − ν02

)|K R |

(1 − ν0

2

)a0 R

2

= (1 − ν0)∣∣Q (a0 R2, R)∣∣,

which contradicts the assumption (3.4). �This lemma shows that at the time level t0, the portion of the cube K R where u(x, t0) is close to its supremum is small.

The second step of the full proof is to bring this information up to all time levels near the top of the cylinder Q (a0 R2, R).

Lemma 2. There exists 1 < s1 ∈ N, depending on the data and ω, such that, for all t ∈ (t0,0),∣∣∣∣x ∈ K R : u(x, t) > μ+ − ω2s1∣∣∣∣ <

(1 −

(ν0

2

)2)|K R |. (3.11)


Proof. Consider the cylinder Q (t0, R) and the level k = μ+ − ω2 . Define

u − k � H+k ≡ ess supQ (t0,R)

(u − k)+ � ω2

,

which we assume to be strictly positive (otherwise there will be nothing to prove). Select n ∈ N big enough so that 0 < c =ω

2n+1 < H+k . Then the logarithmic function ψ

+ given by

ψ+ ={

ln(H+k

H+k −u+k+c) if u > k + c,

0 if u � k + cis well defined and satisfies the inequalities

ψ+ � ln(2n

) = n ln 2, since H+kH+k − u + k + c

�H+kc

�ω2ω

2n+1= 2n;

and, for u = k + c,

0 �(ψ+

)′ � 1H+k − u + k + c

� 1c

= 2n+1

ω

and (ψ+

)′′ = [(ψ+)′]2 � 0.In the weak formulation (2.2) consider the integration over K R × (t0, t), where t ∈ (t0,0), and take φ = 2ψ+(ψ+)′ξ2,

being x �→ ξ(x) a smooth cutoff function defined in K R that verifies⎧⎪⎪⎨⎪⎪⎩

0 � ξ � 1 in K R;ξ ≡ 1 in K(1−σ )R , for some σ ∈ (0,1);|∇ξ | � 1

σ R.

Then, for all t ∈ (t0,0),

J1 + J2 :=t∫

t0

∫K R

ut2ψ+(ψ+)′ξ2 + ∑

t∫t0

∫K R

(umi

)xi

(2ψ+

(ψ+

)′ξ2

)xi

= 0.Using the estimate of ψ+ and Lemma 1, one gets the following estimate for J1

J1 =∫

K R×{t}

(ψ+

)2ξ2 −

∫K R×{t0}

(ψ+

)2ξ2

�∫

K R×{t}

(ψ+

)2ξ2 − n2 ln2 2∣∣x ∈ K R : u(x, t0) > k + c∣∣

�∫

K R×{t}

(ψ+

)2ξ2 − n2 ln2 2

(1 − ν01 − ν02

)|K R |;

as for J2 we get the following estimates using Cauchy’s inequality, the estimates of |∇ξ | and ψ+ , recalling what −t0 � a0 R2and that u � 1

J2 =∑ t∫

t0

∫K R

miumi−1|uxi |22

(1 + ψ+)[(ψ+)′]2ξ2 + ∑

t∫t0

∫K R

miumi−1uxi 2ψ+

(ψ+

)′2ξξxi

�∑ t∫ ∫

miumi−1|uxi |22

[(ψ+

)′]2ξ2 −

∑ t∫ ∫miu

mi−12ψ+|ξxi |2

t0 K R t0 K R


� −∑ t∫

t0

∫K R

miumi−12ψ+|ξxi |2

� −2mn ln 2 1σ 2

a0|K R |.We then get, for all t ∈ (t0,0),∫

K R×{t}

(ψ+

)2ξ2 �

(n2 ln2 2

(1 − ν01 − ν02

)+ 2mn ln 2 1

σ 2a0

)|K R |. (3.12)

Consider the set

S ={

x ∈ K(1−σ )R : u(x, t) > μ+ − ω2n+1

}.

On S , ξ ≡ 1 and ψ+ � (n − 1) ln 2, becauseH+k

H+k − u + k + c�

H+kH+k − ω2 + ω2n

= H+k − ω2 + ω2

H+k − ω2 + ω2n� 2n−1,

since one has H+k − ω2 � 0 and ω2 > ω2n , ∀n > 1. Therefore

(n − 1)2 ln2 2|S| �∫

S×{t}

(ψ+

)2ξ2 �

∫K R×{t}

(ψ+

)2ξ2.

Consequently, for all t ∈ (t0,0),∣∣∣∣x ∈ K R : u(x, t) > μ+ − ω2n+1∣∣∣∣ � |S| + Nσ |K R |�

{(n

n − 1)2( 1 − ν0

1 − ν02

)+ 2ma0

ln 2σ 2n+ Nσ

}|K R |.

Choosing σ so small that Nσ � 38 ν20 and then n so large that

2ma0ln 2σ 2

1

n� 3

8ν20 and

(n

n − 1)2

�(

1 − ν02

)(1 + ν0) ≡ β > 1,

the proof is complete by taking the number s1 = n + 1. �Remark 5. Note that, from the choice of σ , we get σ � 38N ν20 and from the two conditions on n we obtain

n � max{

C(N,m)ω1−mν−60 ;4

ν20+ 2

}.

Recalling that ν0, defined in (3.7), depends on the data and on ω, the same holds for s1.

From this lemma we obtain

Corollary 2. There exists 1 < s1 ∈ N, depending on the data and ω, such that, for all t ∈ (− ν02 a0 R2,0),∣∣∣∣x ∈ K R : u(x, t) > μ+ − ω2s1∣∣∣∣ <

(1 −

(ν0

2

)2)|K R |. (3.13)

From Corollary 2 we deduce that, within the cylinder Q ( ν02 a0 R2, R), the set where u is close to its supremum can be

made arbitrarily small.

Lemma 3. For all ν ∈ (0,1) there exists s1 < s2 ∈ N, depending on the data and on ω, such that∣∣∣∣(x, t) ∈ Q(

ν0

2a0 R

2, R

): u(x, t) > μ+ − ω

2s2

∣∣∣∣ � ν∣∣∣∣Q

(ν0

2a0 R

2, R

)∣∣∣∣.


Proof. Consider the cylinder Q (ν0a0 R2,2R) and the levels k = μ+ − ω2s , for s � s1. Taking in (2.2) φ = (u − k)+ξ2, where0 � ξ � 1 is a smooth cutoff function defined in Q (ν0a0 R2,2R) and satisfying⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

ξ ≡ 0 on ∂p Q(ν0a0 R

2,2R);

ξ ≡ 1 in Q(

ν0

2a0 R

2, R

);

|∇ξ | � 1R

, 0 � ξt �1

ν02 a0 R

2,

we arrive at, for t ∈ (−ν0a0 R2,0),t∫

−ν0a0 R2

∫K2R

ut(u − k)+ξ2 +∑ t∫

−ν0a0 R2

∫K2R

miumi−1uxi

((u − k)+ξ2

)xi

= 0.

Estimating the two integrals

t∫−ν0a0 R2

∫K2R

ut(u − k)+ξ2 = 12

∫K2R×{t}

(u − k)2+ξ2 −t∫

−ν0a0 R2

∫K2R

(u − k)2+ξξt

� −(

ω

2s

)2 1ν02 a0 R

2

t∫−ν0a0 R2

∫K2R

χ[u>k]

� −(

ω

2s

)2 2N+1ν02 a0 R

2

∣∣∣∣Q(

ν0

2a0 R

2, R

)∣∣∣∣,since (u − k)+ � ω2s ;

∑ t∫−ν0 R2

∫K2R

miumi−1uxi

((u − k)+ξ2

)xi

� 12

1

a0

t∫−ν0 R2

∫K R

∣∣∇(u − k)+∣∣2ξ2

− 2m(

ω

2s

)2 2N+1R2

∣∣∣∣Q(

ν0

2a0 R

2, R

)∣∣∣∣,by means of Cauchy’s inequality with = 12 , and observing that when (u − k)+ is not zero

ω

4< μ+ − ω

2� μ+ − ω

2s1� μ+ − ω

2s= k < u � 1,

1

a0<

(ω

4

)mi−1� umi−1 � 1.

Combining the above estimates we arrive at

0∫− ν02 a0 R2

∫K R

∣∣∇(u − k)+∣∣2 �(

ω

2s

)2 2N+3R2

(1

ν0+ ma0

)∣∣∣∣Q(

ν0

2a0 R

2, R

)∣∣∣∣.

We now consider the levels l = μ+ − ω2s+1 > k = μ+ − ω2s , s = s1, . . . , s2 − 1, and define, for t ∈ (− ν02 a0 R2,0),

As(t) ≡{

x ∈ K R : u(x, t) > μ+ − ω2s

}

and

As ≡0∫

− ν0 a R2

∣∣As(t)∣∣dt.

2 0


By Lemma 2.2 and Remarks 2.2 and 2.3 of [6, p. 5] applied to the function u(·, t) for all times t ∈ (− ν02 a0 R2,0),(ω

2s+1

)∣∣As+1(t)∣∣ � C(N) RN+1|K R \ As(t)|∫

[k


I1 + I2 :=t∫

− ν02 a0 R2n

∫K Rn

ut((uω − kn)+ξ2n

) + ∑t∫

− ν02 a0 R2n

∫K Rn

miumi−1uxi

((uω − kn)+ξ2n

)xi

= 0.Reasoning as in the proof of Proposition 2 we obtain

sup− ν02 a0 R2n


Proof. From the previous result we have

u(x, t) � μ+ − ω2s2+1

a.e. (x, t) ∈ Q(

ν0

2

(R

2

)2,√

a0R

2

).

Then

ess oscQ (

ν02 (

R2 )

2,√

a0R2 )

u = ess supQ (

ν02 (

R2 )

2,√

a0R2 )

u − ess infQ (

ν02 (

R2 )

2,√

a0R2 )

u � μ+ − ω2s2+1

− μ−

=(

1 − 12s2+1

)ω =: σ1ω. �

3.3. Proof of Theorem 1

The proof of Proposition 1 is a straightforward consequence of the reduction of the oscillation of u obtained from theanalysis of the first and the second alternatives. If we take

σ = max{σ0,σ1} = σ1, since s2 > 1,and, since ν0 ∈ (0,1),

Q

(ν0

2a0

(R

2

)2,

R

2

)⊂ Q

(a0

(R

2

)2,

R

2

),

at once one gets

ess oscQ (

ν02 a0(

R2 )

2, R2 )

u � σ(ω)ω.

In what follows we show how we can recursively define two sequences of positive real numbers corresponding to radiusand oscillations. Let

ω1 = σ(ω)ω and R1 = RC(ω)

,

where C(ω) = 8ν0σ(ω)m−1

> 8. Defining

Q 1 = Q(a1 R

21, R1

), a1 =

(ω1

4

)1−mobserve that

a1 R21 =

ν0

2a0

(R

2

)2and R1 <

R

8,

then

Q 1 ⊂ Q(

ν0

2a0

(R

2

)2,

R

2

)and ess osc

Q 1u � ess osc

(ν02 a0(

R2 )

2, R2 )

u � σ(ω) = ω1.

This procedure can now be repeated starting from the cylinder Q 1.The sequences mentioned above are defined by

{ω0 = ω,ωn+1 = σ(ωn)ωn and

⎧⎨⎩

R0 = R,Rn+1 = Rn

C(ωn)

and the cylinders are

Q 0 = Q(a0 R

2, R), Q n =

(an R

2n, Rn

), an =

(ωn

4

)1−m.

We will prove Theorem 1 by showing that the sequences (ωn)n and (Rn)n are decreasing sequences converging to zero;moreover, for n = 0,1,2, . . . ,

Q n+1 ⊂ Q n and ess osc u � ωn. (3.16)

Q n


It’s easy to verify that (ωn)n and (Rn)n are decreasing sequences and that they are bounded from below by zero. In orderto guarantee that they really converge to zero, we will prove that they cannot converge to a positive number. Regarding thesequence (Rn)n , since

Rn+1Rn

<1

8

this is easily verified; as for the sequence (ωn)n , the proof is not that simple. Note that

ωn+1ωn

= σ(ωn) ↗ 1,and assume that ωn ↘ α > 0. Therefore

σ(ωn) ↗ σ(α) < 1 and ωn+1 = σ(ωn)ωn � σ(α)ωn, n = 0,1, . . . ,and thereby

ωn+1 � σ n(α)ω, n = 0,1, . . . .Then ωn → 0, contradicting the hypothesis.

The second part is just a consequence of the iterative procedure used to define the sequences.

4. The singular equation

In this section we consider Eq. (2.1) under the assumption 0 < m < 1, that is, we assume that the parabolic pde issingular in its principal part: the diffusion coefficient miumi−1 explodes at the points where u = 0. In this case, in orderto accommodate the singularity, one has to work within cylinders that are stretched in the x-direction. In this context theproper cylinder, as it can be checked by performing some calculations, is

Q(

R2,√

a0 R), a0 =

(ω

4

)1−m.

Remark 6. For the sake of completeness we present all the results concerning the singular case, however in the proofs wedecided to include only the estimates that really depend on the singular character of (2.1).

As we did for the degenerate case, and it is now a standard procedure, we analyze an alternative argument from eachwe will obtain the reduction of the oscillation of u within the smaller cylinder Q ( ν02 (

R2 )

2,√

a0R2 ). The local continuity of u

is then an immediate consequence.

4.1. The alternative reasoning

Assume that μ− = 0. This is a reasonable assumption since 0 � u � 1 and Eq. (2.1) is singular at the points where u = 0.Given ν0 ∈ (0,1), to be determined in terms of the data, either

the first alternative∣∣∣∣(x, t) ∈ Q (R2,√a0 R): u(x, t) < ω2∣∣∣∣ � ν0∣∣Q (R2,√a0 R)∣∣ (4.1)

or

the second alternative∣∣∣∣(x, t) ∈ Q (R2,√a0 R): u(x, t) > μ+ − ω2∣∣∣∣ < (1 − ν0)∣∣Q (R2,√a0 R)∣∣. (4.2)

We start by assuming that the first inequality holds. Then

Proposition 4. There exists ν0 ∈ (0,1), depending only on the data, such that if (4.1) holds true then

u(x, t) � ω4

, a.e. (x, t) ∈ Q((

R

2

)2,√

a0R

2

). (4.3)


Proof. Define two decreasing sequences of positive numbers

Rn = R2

+ R2n+1

, kn = ω4

+ ω2n+2

, n = 0,1, . . .and construct the family of nested and shrinking cylinders Q (R2n,

√a0 Rn).

Introduce the function uω = max{u, ω4 } and, in the weak formulation (2.2), take φ = −(uω − kn)−ξ2n , where 0 � ξn � 1are smooth cutoff functions defined in Q n = Q (R2n,

√a0 Rn) and satisfying⎧⎪⎨

⎪⎩ξn ≡ 1 in Q n+1, ξn ≡ 0 on ∂p Q n;

|∇ξn| � 2n+2

√a0 R

, |�ξn| � 22(n+2)

a0 R2, 0 < ∂tξn �

22(n+2)

R2,

and consider the integration in time to be held over (−R2n, t), for t ∈ (−R2n,0). We obtain

I1 + I2 :=t∫

−R2n

∫K√a0 Rn

ut[−(uω − kn)−ξ2n ] + ∑

t∫−R2n

∫K√a0 Rn

(umi

)xi

[−(uω − kn)−ξ2n ]xi= 0.

Arguing as in the proof of Proposition 2

I1 �1

2

∫K√a0 Rn ×{t}

(uω − kn)2−ξ2n − 3(

ω

4

)2 22(n+2)R2

t∫−R2n

∫√

a0 K Rn

χ[uω�kn];

concerning I2

I2 =∑ t∫

−R2n

∫K√a0 Rn

miumi−1ω

∣∣((uω − kn)−)xi ∣∣2ξ2n + 2∑t∫

−R2n

∫K√a0 Rn

miumi−1ω

((uω − kn)−

)xiξn(uω − kn)−ξnxi

+ 2(

ω

2n+2

)∑ t∫−R2n

∫K√a0 Rn

( ω4∫u

mi smi−1 ds

)xi

ξnξnxi χ[u< ω4 ]

� 12

m

t∫−R2n

∫K√a0 Rn

∣∣∇(uω − kn)−∣∣2ξ2n − 2ma0(

ω

4

)2 22(n+2)a0 R2

t∫−R2n

∫K√a0 Rn

χ [uω < kn]

− 4(

ω

4

)2(ω

4

)m−1 22(n+2)a0 R2

t∫−R2n

∫K Rn

χ[u< ω4 ]

= 12

m

t∫−R2n

∫K√a0 Rn

∣∣∇(uω − kn)−∣∣2ξ2n − (2m + 4)(

ω

4

)2 22(n+2)R2

t∫−R2n

∫K√a0 Rn

χ [uω < kn].

The above inequalities were obtained by Cauchy’s inequality with = 12 , integration by parts, the fact that for ω4 < u =uω � kn , we get

(uω − kn)− = kn − uω = kn − u < ω2n+2

� ω4

;

1 < umi−1ω �(

4

ω

)1−mi< a0

and for u � ω4ω4∫

mi smi−1 ds =

(ω

4

)mi− umi �

(ω

4

)mi�

(ω

4

)m.

u


Combining the estimates obtained for I1 and I2 we arrive at

sup−R2n


Corollary 3. There exist constants ν0, σ0 ∈ (0,1), depending only on the data, such that if (3.3) holds theness osc

Q (( R2 )2,

√a0

R2 )

u � σ0ω. (4.5)

Now we assume that the first alternative does not hold, and therefore the second alternative must be in force.Being the number ν0 already determined we will show, by performing several proofs in between, that even in this case

one can get a reduction of the oscillation of u in a full cylinder smaller than Q (R2,√

a0 R) but with the same vertex.

Lemma 5. Assume that (3.4) holds true. There exists a time level

t0 ∈[−R2,−ν0

2R2

](4.6)

such that∣∣∣∣x ∈ K√a0 R : u(x, t0) > μ+ − ω2∣∣∣∣ <

(1 − ν01 − ν02

)|K√a0 R |. (4.7)

Remark 7. We omit the proof since it does not give any new information/estimate regarding the corresponding one per-formed in the degenerate case.

This lemma shows that at the time level t0, the portion of the cube K R where u(x, t0) is close to its supremum is small.Our aim is to bring this information up to all levels near the top of the cylinder Q (R2,

√a0 R). This will be accomplished in

two steps, being the first one

Lemma 6. There exists 1 < s1 ∈ N, depending only on the data, such that, for all t ∈ (t0,0),∣∣∣∣x ∈ K√a0 R : u(x, t) > μ+ − ω2s1∣∣∣∣ <

(1 −

(ν0

2

)2)|K√a0 R |. (4.8)

Proof. Consider the cylinder Q (t0,√

a0 R) and the level k = μ+ − ω2 . Define

u − k � H+k ≡ ess supQ (t0,R)

(u − k)+ � ω2

,

which we assume to be strictly positive (otherwise there will be nothing to prove). Select n ∈ N big enough so that 0 < c =ω

2n+1 < H+k . Then the logarithmic function ψ

+ given by

ψ+ ={

ln(H+k

H+k −u+k+c) if u > k + c,

0 if u � k + cis well defined and satisfies the inequalities

ψ+ � ln(2n

) = n ln 2,and, for u = k + c,

0 �(ψ+

)′ � 1H+k − u + k + c

� 1c

= 2n+1

ω

and (ψ+

)′′ = [(ψ+)′]2 � 0.In the weak formulation (2.2) consider the integration over K√a0 R × (t0, t), where t ∈ (t0,0), and take φ = 2ψ+(ψ+)′ξ2,

being x �→ ξ(x) a smooth cutoff function defined in K R that verifies⎧⎪⎪⎪⎨⎪⎪⎪⎩

0 � ξ � 1 in K√a0 R;ξ ≡ 1 in K(1−σ )√a0 R , for some σ ∈ (0,1);

|∇ξ | � 1√ .

σ a0 R


Then, for all t ∈ (t0,0),

J1 + J2 :=t∫

t0

∫K√a0 R

ut2ψ+(ψ+)′ξ2 + ∑

t∫t0

∫K√a0 R

(umi

)xi

(2ψ+

(ψ+

)′ξ2

)xi

= 0.Using the estimate for ψ+ and Lemma 1 one gets the following estimate for J1

J1 =∫

K√a0 R×{t}

(ψ+

)2ξ2 −

∫K√a0 R×{t0}

(ψ+

)2ξ2

�∫

K√a0 R×{t}

(ψ+

)2ξ2 − n2 ln2 2∣∣x ∈ K√a0 R : u(x, t0) > k + c∣∣

�∫

K√a0 R×{t}

(ψ+

)2ξ2 − n2 ln2 2

(1 − ν01 − ν02

)|K√a0 R |,

analogously to what we have done in the degenerate case. Concerning J2 we get the following estimates by using Cauchy’sinequality, the estimates for |∇ξ | and ψ+ , recalling that −t0 � R2 and that, since ψ+ is not zero at the points whereu > k + c = ω2 + ω2n+1 � ω4 , one has umi−1 � a0,

J2 =∑ t∫

t0

∫K√a0 R

miumi−1|uxi |22

(1 + ψ+)[(ψ+)′]2ξ2 + ∑

t∫t0

∫K√a0 R

miumi−1uxi 2ψ+

(ψ+

)′2ξξxi

�∑ t∫

t0

∫K√a0 R

miumi−1|uxi |22

[(ψ+

)′]2ξ2 −

∑ t∫t0

∫K√a0 R


� −∑ t∫

t0

∫K√a0 R


� −2mn ln 2 1σ 2

|K√a0 R |.For all t ∈ (t0,0), we obtain∫

K√a0 R×{t}

(ψ+

)2ξ2 �

(n2 ln2 2

(1 − ν01 − ν02

)+ 2mn ln 2 1

σ 2

)|K√a0 R |. (4.9)

Consider the set

S ={

x ∈ K(1−σ )√a0 R : u(x, t) > μ+ −ω

2n+1

}.

On S , ξ ≡ 1 and ψ+ � (n − 1) ln 2; therefore

(n − 1)2 ln2 2|S| �∫

S×{t}

(ψ+

)2ξ2 �

∫K√a0 R×{t}

(ψ+

)2ξ2.

Consequently, for all t ∈ (t0,0),∣∣∣∣x ∈ K√a0 R : u(x, t) > μ+ − ω2n+1∣∣∣∣ � |S| + Nσ |K R |�

{(n

n − 1)2( 1 − ν0

1 − ν0)

+ mln 2σ 2n

+ Nσ}|K√a0 R |.

2


Choosing σ so small that Nσ � 38 ν20 and then n so large that

m

ln 2σ 21

n� 3

8ν20 and

(n

n − 1)2

�(

1 − ν02

)(1 + ν0) ≡ β > 1,

the proof is complete by taking the number s1 = n + 1. �Remark 8. Clearly the number s1 does not dependent on ω due to the choice of ν0 performed in (4.4).

From this lemma we obtain

Corollary 4. There exists 1 < s1 ∈ N, depending only on the data, such that, for all t ∈ (− ν02 R2,0),∣∣∣∣x ∈ K√a0 R : u(x, t) > μ+ − ω2s1∣∣∣∣ <

(1 −

(ν0

2

)2)|K√a0 R |. (4.10)

From this result we deduce that, within the cylinder Q ( ν02 R2,

√a0 R), the set where u is close to its supremum can be

made arbitrarily small.

Lemma 7. For all ν ∈ (0,1) there exists s1 < s2 ∈ N, depending on the data and on ω, such that∣∣∣∣(x, t) ∈ Q(

ν0

2R2,

√a0 R

): u(x, t) > μ+ − ω

2s2

∣∣∣∣ � ν∣∣∣∣Q

(ν0

2R2,

√a0 R

)∣∣∣∣.Proof. Consider the cylinder Q (ν0 R2,2

√a0 R) and the levels k = μ+ − ω2s , for s � s1. Taking in (2.2) φ = (u − k)+ξ2, where

0 � ξ � 1 is a smooth cutoff function defined in Q (ν0 R2,2√

a0 R) and satisfying⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

ξ ≡ 0 on ∂p Q(ν0 R

2,2√

a0 R);

ξ ≡ 1 in Q(

ν0

2R2,

√a0 R

);

|∇ξ | � 1√a0 R

, 0 � ξt �1

ν02 R

2,

we arrive at, for t ∈ (−ν0 R2,0)t∫

−ν0 R2

∫K2√a0 R

ut(u − k)+ξ2 +∑ t∫

−ν0 R2

∫K2√a0 R

miumi−1uxi

((u − k)+ξ2

)xi

= 0.

To estimate the integral evolving the time derivative we proceed as in the proof of Lemma 3, and then get

t∫−ν0 R2

∫K2√a0 R

ut(u − k)+ξ2 � 12

∫K2√a0 R×{t}

(u − k)2+ξ2 −(

ω

2s

)2 2N+1ν02 R

2

∣∣∣∣Q(

ν0

2R2,

√a0 R

)∣∣∣∣,since (u − k)+ � ω2s .

To estimate the integral evolving the space derivatives we perform as follows

∑ t∫−ν0 R2

∫K2√a0 R

miumi uxi

((u − k)+ξ2

)xi

� m2

t∫−ν0 R2

∫K√a0 R

∣∣∇(u − k)+∣∣2ξ2 − 2m(

ω

2s

)2 2N+1R2

∣∣∣∣Q(

ν0

2R2,

√a0 R

)∣∣∣∣,by means of Cauchy’s inequality with = 12 , and observing that when (u − k)+ is not zero

ω

4< μ+ − ω

2� μ+ − ω

2s1� μ+ − ω

2s= k < u � 1, 1 � umi−1 � a0.

Combining the above estimates we arrive at

0∫− ν0 R2

∫K√a R

∣∣∇(u − k)+∣∣2 � 1mν0

(ω

2s

)2 2N+4R2

∣∣∣∣Q(

ν0

2R2,

√a0 R

)∣∣∣∣.

2 0


We now consider the levels l = μ+ − ω2s+1 > k = μ+ − ω2s , s = s1, . . . , s2 − 1, and define, for t ∈ (− ν02 R2,0),

As(t) ≡{

x ∈ K√a0 R : u(x, t) > μ+ −ω

2s

}

and

As ≡0∫

− ν02 R2

∣∣As(t)∣∣dt.

By Lemma 2.2 and Remarks 2.2 and 2.3 of [6, p. 5] applied to the function u(·, t) for all times t ∈ (− ν02 R2,0),(ω

2s+1

)∣∣As+1(t)∣∣ � C(N) (√

a0 R)N+1

|K R \ As(t)|∫

[k


and construct the family of nested and shrinking cylinders

Q n = Q(

ν0

2R2n,

√a0 Rn

).

Consider the function uω = min{u,μ+ − ω2s2+1 } and, in the weak formulation (2.2), take φ = (uω − kn)+ξ2n , where 0 �ξn � 1 are smooth cutoff functions defined in Q n and verifying⎧⎪⎨

⎪⎩ξn ≡ 1 in Q n+1, ξn ≡ 0 on ∂p Q n;

|∇ξn| � 2n+2

√a0 R

, |�ξn| � 22(n+2)

a0 R2, 0 < ∂tξn �

22(n+2)ν02 R

2.

For t ∈ (− ν02 R2n,0), we arrive at

I1 + I2 :=t∫

− ν02 R2n

∫K√a0 Rn

ut((uω − kn)+ξ2n

) + ∑t∫

− ν02 R2n

∫K√a0 Rn

miumi−1uxi

((uω − kn)+ξ2n

)xi

= 0.Reasoning as in the proof of Proposition 4 we obtain

sup− ν02 R2n


which implies Y0 � ν . Therefore, using the fast geometric convergence result Lemma 4.1 of [6, p. 12], Yn → 0 when n → ∞,which proves the required. �

Now we are able to present and prove the result concerning the reduction of the oscillation. Namely,

Proposition 5. There exist positive numbers ν0 , depending only on the data and σ1 ∈ (0,1), depending on the data and on ω, suchthat, if (3.4) holds true then

ess oscQ (

ν02 (

R2 )

2,√

a0R2 )

u � σ1ω. (4.12)

Proof. The proof is similar to the proof of Corollary 1; we just have to take σ1 = 1 − 12s2+1 . The dependence on ω comesvia s2. �Remark 10. The proof of Theorem 1 is analogue to the one presented for the degenerate case.

The dependence on ω occurs in different forms when comparing the degenerate case and the singular one. As for thedegenerate case, the dependence on ω appears almost “instantaneously”, when we are determining the first constant ν0;in the singular case, everything works just fine till almost the end of the study of the alternatives, the first and uniqueconstant depending on ω is s2.

References

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Concerning the regularity of the anisotropic porous medium equationIntroductionRegularity resultThe degenerate equationReduction of the oscillation when u is essentially away from zeroReduction of the oscillation when u is essentially away from its supremumProof of Theorem 1

The singular equationThe alternative reasoning

References

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